Evaluate the following integrals: _ 0 ^ 2 1 5 +3 dx

Question

Evaluate the following integrals:
$\int_{0}^\limits{\frac{\pi}{2}}\frac{1}{5\cos\text{x}+3\sin\text{x}}\text{ dx}$

✓

Answer

Let $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{1}{5\cos\text{x}+3\sin\text{x}}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{1}{5\Bigg(\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)+3\Bigg(\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)}\text{ dx}$ $\Bigg[\because\sin\text{A}=\bigg(\frac{2\tan\frac{\text{A}}{2}}{1+\tan^2\frac{\text{A}}{2}}\bigg),\cos\text{A}=\bigg(\frac{1-\tan^2\frac{\text{A}}{2}}{1+\tan^2\frac{\text{A}}{2}}\bigg)\Bigg]$
$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{1+\tan^2\frac{\text{x}}{2}}{5-5\tan^2\frac{\text{x}}{2}+6\tan\frac{\text{x}}{2}}\text{ dx}$
$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{\sec^2\frac{\text{x}}{2}}{5-5\tan^2\frac{\text{x}}{2}+6\tan\frac{\text{x}}{2}}\text{ dx}$
Let $\tan\frac{\text{x}}{2}=\text{t}$ Then, $\frac{1}{2}\sec^2\frac{\text{x}}{2}\text{ dx}=\text{dt}$
Also, $\text{x}=0,\text{t}=0$ and $\text{x}=\frac{\pi}{2}, \text{t}=1$
$\therefore\ \text{I}=\int_{0}^\limits{1}\frac{2\text{ dt}}{5-5\text{t}^2+6\text{t}}$
$\text{I}=\frac{1}{5}\int_{0}^\limits{1}\frac{2\text{dt}}{1-\text{t}^2+\frac{6}{5}\text{t}+\frac{36}{100}-\frac{36}{100}}$
$=\frac{2}{5}\int_{0}^\limits{1}\frac{\text{dt}}{-\big(\text{t}-\frac{6}{10}\big)^2+\frac{136}{100}}$
$=\frac{2}{5}\times\frac{10}{\sqrt{136}}\Bigg[-\log\Bigg(\frac{\text{t}-\frac{6}{10}-\frac{\sqrt{136}}{10}}{\text{t}-\frac{6}{10}+\frac{\sqrt{136}}{10}}\Bigg)\Bigg]^1_0$
$=\frac{1}{\sqrt{34}}\bigg[-\log\bigg(\frac{4-2\sqrt{34}}{4+2\sqrt{34}}\bigg)+\log\bigg(\frac{-6-2\sqrt{34}}{-6+2\sqrt{34}}\bigg)\bigg]$
$=\frac{1}{\sqrt{34}}\log\bigg(\frac{6+2\sqrt{34}}{6-2\sqrt{34}}\times\frac{4+2\sqrt{34}}{4-2\sqrt{34}}\bigg)$
$=\frac{1}{\sqrt{34}}\log\bigg(\frac{160+20\sqrt{34}}{160-20\sqrt{34}}\bigg)$
$=\frac{1}{\sqrt{34}}\log\bigg(\frac{8+\sqrt{34}}{8-\sqrt{34}}\bigg)$

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